# How many liters of a .88 M solution can be made with 25.5 grams of lithium fluoride (LiF)?

May 16, 2016

$L = 1.10 L$ of solution

#### Explanation:

The Molarity $M$ is calculated by the equation comparing moles of solute to liters of solution

$M = \frac{m o l}{L}$

For this question we are given the Molarity 0.88M

We are told the solute is a 25.2 gram sample of LiF, Lithium Fluoride

We can convert the mass of LiF to moles by dividing by the molar mass of LiF

Li = 6.94
F = 19.0

LiF = 25.94 g/mole

$25.2 \cancel{g r a m s} x \frac{1 m o l}{25.94 \cancel{g r a m s}}$ = $0.97$ moles

Now we can take the the molarity and the moles and calculate the Liters of solution

$M = \frac{m o l}{L}$

$M L = m o l$

$L = \frac{m o l}{M}$

$L = \frac{0.97 m o l}{0.88 M}$

$L = 1.10 L$ of solution